Let $\lambda$ be an ordinal. A cardinal $\kappa$ is $\lambda$-strong iff there is some inner model $M$ and an elementary embedding $$ j \colon V \rightarrow M $$ s.t. $crit(j) = \kappa$ and $V_\lambda \subseteq M$.
I am given the following exercise:
Let $\alpha < \beta < \kappa$ be ordinals, where $\kappa$ is $(\kappa+\beta)$-strong. Then $$ \{ \mu < \kappa \mid \mu \text{ is } (\kappa+\alpha) \text{-strong} \} $$ has size $\kappa$.
I'd like to show that $$ M \models \kappa \text{ is } (j(\kappa)+\alpha) \text{-strong}, $$ but I'm not sure where to begin...
(It seems that I have to "stretch" a $(\kappa, \kappa+\alpha)$-extender $E \in V$ to a $(\kappa, j(\kappa)+\alpha)$-extender $\tilde E \in M$, but how? Is this even the right approach?)
edit:
Consider the case $\alpha =0 $ and recall that a cardinal $\mu$ is $\mu$-strong iff it is measurable.
Now, if $\kappa$ is $\kappa+2$-strong, we may fix a witnessing measure $U$ on $\kappa$. Then $U \in V_{\kappa+2} \subseteq M$ and some basic observations yield $$ M \models \kappa \text{ is measurable} $$ Thus for any fixed $\xi < \kappa$ we get $$ M \models \exists \mu \colon j(\xi) < \mu < j(\kappa) \wedge \mu \text{ is measurable} $$ (take $\mu = \kappa$) and elementarity yields that there is some measurable $\mu \in (\xi, \kappa)$. This proves that there are $\kappa$ many $\mu$-strong cardinals $\mu < \kappa$ and also suggest that we should be more careful and require something like $\alpha+n < \beta < \kappa$ for some $0 < n < \omega$. (Given the kind of proof I have for the edited question below, the exact value of $n$ is then easily calculated.)
I think that this exercise contains a typo and should read $$ \{ \mu < \kappa \mid \mu \text{ is } (\mu+\alpha) \text{-strong} \} $$ instead. I will ask my professor and in the case that this is what he was asking for, I will tidy up my post and provide an answer for future reference.